Group Cohomology as a Derived Functor — Examples

Degree zero cochains are $M$, while every positive normalized bar is zero because its only group entry is $1$.

The normalized complex is $M$ in degree zero and zero above it, proving the computation.

Fixed points satisfy $m=-m$, hence $m=0$ in $\mathbb{Z}$.

Coinvariants quotient by $tm-m=-2m$, hence are $\mathbb{Z}\mathrm{/}2$; degree-zero recovery identifies these with the two asserted groups.

Substitution in the definition yields the two displayed formulas and $d(g_0)=0$.

Applying $d$ to the three terms of the second formula produces each vertex twice with opposite signs, explicitly illustrating $d^2=0$.

Set $g=h=1$ to get $z(1)=z(1)+z(1)$.

Subtracting $z(1)$ gives $z(1)=0$. Thus the normalized representative is $z$ itself.

$(t-1)N=N(t-1)=t^m-1=0$, so this is a complex.

Writing an element as $\sum a_i{t}^i$ shows $\ker \epsilon =(t-1)\mathbb{Z}[C_m]$, $\ker (t-1)=N\mathbb{Z}[C_m]$, and $\ker N=(t-1)\mathbb{Z}[C_m]$. Hence it is exact and every term is free.

Applying ${\mathrm{Hom}}_{\mathbb{Z}[C_m]}(-,M)$ identifies every cochain group with $M$; the coboundaries alternate between $t-1$ and $N$.

Taking kernel modulo preceding image gives exactly the displayed groups, including the degree-zero kernel.

Induction and coinduction from $1$ are respectively $\mathbb{Z}[G]\otimes M$ and ${\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z}[G],M)$.

The assumed Choice supplies the transversal premise used in the two cited Shapiro theorems. Shapiro identifies their positive (co)homology with positive (co)homology of the trivial group, which vanishes.

The contraction has $s(1)=(1,1)$, so $hs(1)=(h,h)$.

But $s(h\cdot (1))=s(h)=(1,h)$, which differs from $(h,h)$. Thus equivariance fails.

The trivial module for $1$ is free, so its projective dimension is zero. For $\mathbb{Z}$, $0\to \mathbb{Z}[t,t^{-1}]\stackrel{t-1}{\to }\mathbb{Z}[t,t^{-1}]\to \mathbb{Z}\to 0$ is a free resolution.

With trivial coefficients $\mathbb{Z}$, applying Hom sends $t-1$ to zero, so $H^1(\mathbb{Z};\mathbb{Z})=\mathbb{Z}\mathrm{\ne }0$. Hence the resolution length one is minimal.